Question

Calculati sumele :

a. 1+2+3+...+2006
b. 2+4+6+...+2006
c. 1+3+5+...+2005

Answer 1

[tex]a)1+2+3+...+2006\\ Aplicam~Suma~Gauss:[n(n+1)}:2\\ \frac{2006(2006+1)}{2} =\ \textgreater \ 2.013.021\\\\ b)2+4+6+...+2006 (dam~factor~comun~pe~2~si~obtinem:)\\ 2(1+2+3+...+1003)\\ Avem~o~Suma~Gauss~si~aplicam~formula:[n(n+1)]:2\\ \frac{1003(1003+1)}{2} *2=1.007.012 [/tex]
[tex]c)1+3+5+...+2005\\ 1=2*0+1\\ 3=2*1+1\\ 5=2*2+1\\ ---------\\ 2005=2*1002+1\\ S=(2*0+1)+(2*1+1)+(2*2+1)+...+(2*1002+1)\\ S=2*0+2*1+2*2+...+2*1002+1+1+1+...+1\\ S=0+2(1+2+3+...+2005)+1003\\ S=0+2*2011015+1003\\ S=0+4022030+1003\\ S=4.023.033[/tex]

Answer 2

a) 1+2+3+...+2006   (se face cu ajutorul sumei Gauss)
=2006*2007:2
=4026042:2
=2013021
b)2+4+6+...+2006  (il dam pe 2 factor comun)
=2(1+2+3+...+1003)
=2*1003*1004:2
=1003*1004
=1007012
c)1+3+5+...+2005
=(2*1-1)+(2*2-1)+...+(2*1003-1)
=1003²
=1006009